In fact n need not divide both a-b and a+b! As an example choose n = 8, a = 3,b=1...

July 8th 2008, 01:20 PM

Vedicmaths

I got until this part:

n divides a^2 - b^2 then we could say for some k, (a-b)(a+b) = nk..now similarly for some l , a+b = nl..

so dividing both of these equations we get: (a-b) = k/l...and I am stuck after this part! is this a right way to proceed??
please help!

July 8th 2008, 01:41 PM

Vedicmaths

a little head up!

well I tried disproving it and got this one. Like you said, just need one example where ti does not work...

For example, a = 7, b = 2, N = 3.

7^2 mod 3 = 1
2^2 mod 3 = 1
so 7^2 mod 3 = 2^2 mod 3

BUT
7 mod 3 = 1
2 mod 3 = 2
so 7 mod 3 != 2 mod 3

So it satisfies our counterexample....but I still don't understand, how do we know that any conditions like this one, proves or disproves? so do we always look for disproving first and pick any examples where it does not work. Or something else?

July 8th 2008, 07:56 PM

Isomorphism

Didnt you follow my reasoning? Its clear from that...

n|a^2-b^2 = (a-b)(a+b)

Now I realised that if n|(a+b) then n|(a^2 - b^2).... This is how I figured n need not divide a-b.